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It means that the boundaries of the set are not included in the set.
For example, consider the set of numbers that are bigger than 1 and smaller than 2. The set is bounded by 1 and 2 but neither of these belong to the set.
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Summation of a set of numbers?
That means to add all the numbers together.
Does the set of irrational numbers contain the set of real numbers?
Yes. If its irrational it just means that it continues forever with no real pattern. It can still have real numbers
Is the image of an open set under a continuous mapping need not be open?
f(x) = x^{2} is a continuous function on the set R of real numbers, and (-1, 1) is an open set in R, but f(-1, 1) = [0, 1), and [0, 1) is not an open set in R. So, f is not an open function on R.
The densest subset of real numbers is the set of fractions?
Your question is ill-posed. I have not come across a comparison dense-denser-densest. The term "dense" is a topological property of a set: A set A is dense in a set B, if for all y in B, there is an open set O of B, such that O and A have nonempty intersection. The rational numbers are indeed dense in the set of real numbers with the standard topology. An open set containing a real number contains always a rational number. Another way of saying it is that every real number can be approximated to any precision by rational numbers. There are denser sets, if you are willing to consider more elements. Suppose you construct a set consisting of the rational numbers plus all algebraic numbers. The set of algebraic numbers is also countable, but adding them, makes it obviously easier to approximate real numbers. Can you perhaps construct a set less dense than the set of rational numbers? Suppose we take the set of rational numbers without the element 0. Is this set still dense in the real numbers? Yes, because 0 can be approximated by 1/n, n>1. In fact, you can remove finite number of rational numbers from the set of rational numbers and the resulting set will still be dense in the set of the real numbers.
What are examples of infinity sets?
Many infinite sets appear in mathematics: the set of counting numbers; the set of integers; the set of rational numbers; the set of irrational numbers; the set of real numbers; the set of complex numbers. Also, certain subsets of these, such as the set of square numbers, the set of prime numbers, and others.
Is the set of real numbers is an open set?
Yes.
What does mean in math means?
The mean is the average of a set of numbers Mean/average = sum of the numbers in the set divided by the amount of numbers in the set
Summation of a set of numbers?
That means to add all the numbers together.
What doen mean mean?
mean means the average or all the numbers in the set added together and then divided by the number of numbers in the set of numbers.
Does the set of irrational numbers contain the set of real numbers?
Yes. If its irrational it just means that it continues forever with no real pattern. It can still have real numbers
Are all numbers whole?
The integers are the set { ...,-3,-2,-1,0,1,2,3,...} where the ... means that the list continues forever. Since this set includes the negative numbers whihc are not whole numbers, the answer would be no. The whole numbers are the set {0,1,2,3,...}
What do you mean by countably infinite and infinite?
Countably infinite means you can set up a one-to-one correspondence between the set in question and the set of natural numbers. It can be shown that no such relationship can be established between the set of real numbers and the natural numbers, thus the set of real numbers is not "countable", but it is infinite.
Are all integers whole numbers?
The integers are the set { ...,-3,-2,-1,0,1,2,3,...} where the ... means that the list continues forever. Since this set includes the negative numbers whihc are not whole numbers, the answer would be no. The whole numbers are the set {0,1,2,3,...}
What is the range of something?
If you mean for a set of numbers in maths then it means the difference between the highest and lowest numbers.
Is the image of an open set under a continuous mapping need not be open?
f(x) = x^{2} is a continuous function on the set R of real numbers, and (-1, 1) is an open set in R, but f(-1, 1) = [0, 1), and [0, 1) is not an open set in R. So, f is not an open function on R.
Are all integers whole?
The integers are the set { ...,-3,-2,-1,0,1,2,3,...} where the ... means that the list continues forever. Since this set includes the negative numbers whihc are not whole numbers, the answer would be no. The whole numbers are the set {0,1,2,3,...}
The densest subset of real numbers is the set of fractions?
Your question is ill-posed. I have not come across a comparison dense-denser-densest. The term "dense" is a topological property of a set: A set A is dense in a set B, if for all y in B, there is an open set O of B, such that O and A have nonempty intersection. The rational numbers are indeed dense in the set of real numbers with the standard topology. An open set containing a real number contains always a rational number. Another way of saying it is that every real number can be approximated to any precision by rational numbers. There are denser sets, if you are willing to consider more elements. Suppose you construct a set consisting of the rational numbers plus all algebraic numbers. The set of algebraic numbers is also countable, but adding them, makes it obviously easier to approximate real numbers. Can you perhaps construct a set less dense than the set of rational numbers? Suppose we take the set of rational numbers without the element 0. Is this set still dense in the real numbers? Yes, because 0 can be approximated by 1/n, n>1. In fact, you can remove finite number of rational numbers from the set of rational numbers and the resulting set will still be dense in the set of the real numbers.
